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Find the general solution of y′′ + y′ – 6y = 2x3 + 5x2 – 7x + 2

by RoRi

Find the general solution of the second-order differential equation

    \[ y'' + y' - 6y = 2x^3 + 5x^2 - 7x + 2. \]

If the solution is not valid everywhere, describe the interval on which it is valid.

The general solution of the homogeneous equation

    \[ y'' + y' - 6y = 0 \]

is given by Theorem 8.7 with a = 1 and b = -6. This gives us d = a^2 - 4b = 25; hence, k = \frac{1}{2} \sqrt{d} = \frac{5}{2}. Thus,

    \begin{align*} y &= e^{-\frac{ax}{2}} (c_1 e^{kx} + c_2 e^{-kx}) \\ &= e^{-\frac{x}{2}} (c_1 e^{\frac{5}{2}x} + c_2 e^{-\frac{5}{2}x}) \\ &= c_1 e^{2x} + c_2 e^{-3x}. \end{align*}

To find a particular solution of y'' + y' - 6y = 2x^3 + 5x^2 - 7x + 2 let y_1 = Ax^3 + Bx^2 + Cx + D. Then,

    \begin{align*} y_1' &= 3Ax^2 + 2Bx + C \\ y_1'' &= 6Ax + 2B. \end{align*}

Therefore,

    \begin{align*} y'' + y' - 6y &= 2x^3 + 5x^2 - 7x + 2 \\ \implies 6Ax + 2B + 3Ax^2 + 2Bx + C - 6Ax^3 - 6Bx^2 -6Cx - 6D &= 2x^3 + 5x^2 - 7x + 2. \end{align*}

Setting the coefficients of like powers of x to be equal and solving for the constants we get

    \[ A = -\frac{1}{3}, \quad B = -1, \quad C = \frac{1}{2}, \quad D = -\frac{7}{12}. \]

By Theorem 8.8 (page 330 of Apostol) the general solution of the given differential equation is then

    \[y = c_1 e^{2x} + c_2 e^{-3x} - \frac{1}{3}x^3 - x^2 + \frac{1}{2}x - \frac{7}{12}. \]

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